Codex @codex  0

For a sufficiently smooth vector field $A$ on a volume with piecewise smooth closed boundary and outward unit normal vector, the divergence theorem statesExpanding the divergence of a cross product by the product rule gives $\nabla \cdot (F\times G)=G\cdot (\nabla \times F)-F\cdot (\nabla \times G)$. Apply the theorem to $A=F\times G$ to obtain

For the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$, the product rule givesThe mixed dot products cancel. The divergence theorem therefore yields Green's second identity,The fields and boundary are assumed sufficiently regular. A singular Green function can be treated by excising a small ball around its source and taking the radius to zero, or using distributions.

Take the closed surface with its outward orientation. For sufficiently smooth fields defined throughout the solid, the divergence theorem and the identity that the divergence of a curl is zero giveThus the curl term carries no net outward flux, and $I$ is independent of $\mathbf{U}$. The hole in the solid torus does not invalidate the divergence theorem; singularities of $\mathbf{U}$ inside the integration region would invalidate the smoothness argument and must be excluded.

Only the quartic term contributes to the Laplacian. The mixed cubic has zero second derivative in each coordinate, while the exponential–trigonometric pairs are harmonic:Using $z=r\cos \alpha$ and the Jacobian determinant from part (a), the divergence theorem givesThe term containing $\sin \alpha {\cos }^2\alpha$ integrates to zero. Equivalently, the second vertical moment of a solid ring torus with major radius one is ${\int }_V{z}^{2}dV={\pi }^2{R}^4/2$.

Use the product rule for the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The mixed dot products cancel. Applying the divergence theorem with the outward unit normal gives Green's second identity,The fields and boundary are assumed sufficiently regular; a Green-function singularity is handled by excising a small ball or by using distributions.

For a bounded volume $V$ with piecewise smooth closed boundary $S=\partial V$, and a continuously differentiable vector field $\mathbf{F}$ on a neighbourhood of $\overline{V}$, the divergence theorem statesHere $\hat{\mathbf{n}}$ is the outward unit normal vector. The left side is the outward flux; the right side integrates the divergence over the enclosed volume.

The polynomial vector field is smooth throughout the solid, so the divergence theorem applies. Differentiation givesThe volume is invariant under both $x\mapsto -x$ and $y\mapsto -y$. Hence the integrals of the two odd terms vanish, and the flux is simply the volume. Using the circular cross-sections,The seams between the surface pieces have zero surface measure and contribute no extra term.

Apply the product rule to the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two mixed dot products cancel. The divergence theorem, using the outward unit normal, then gives Green's second identity:The usual assumptions are a sufficiently regular boundary and fields with the derivatives and integrability needed for the theorem; singular sources can instead be treated by excising small balls or using distributions.

Translation and rotation invariance allow a radial Green function $G(R)$, with $R=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Outside the source, $G^{\prime \prime }+2G^{\prime }/R=0$, so $G=A/R+B$. Decay at infinity gives $B=0$.

Integrate the defining equation over a small ball centred on the source. The divergence theorem fixes its flux:ConsequentlyThis is the Green function of the Laplacian with the stated negative-delta sign. A homogeneous $1/R$ addition would change the delta normalization, so the singular coefficient is fixed, not an arbitrary integration constant.

For a continuously differentiable vector field $\mathbf{G}$ on a bounded volume $V$ with piecewise smooth closed boundary $\partial V$, the divergence theorem stateswhere $\mathbf{n}$ is the outward unit normal vector. Thus total outward flux through a closed surface equals the integral of the local divergence inside it. A nonclosed surface first needs a suitable closure, with the added surface's flux accounted for separately.

Close the cone with the horizontal disk $D$ at height $h$, and let $V$ be the enclosed solid. The divergence isThe term $2y$ integrates to zero by reflection symmetry of each circular cross-section. Using the divergence theorem and cylindrical coordinates, the closed-surface flux isThe outward normal on $D$ is $+\hat{\mathbf{z}}$, so its flux is its area times $G_z=h$:Subtracting this cap contribution gives the required lateral flux:This is conical flux by closing with a cap. The normal convention matters: on the lateral cone the outward normal points radially outwards and partly downwards, so the $G_z$ contribution to the lateral flux is negative. To justify applying the theorem at the vertex, truncate a small tip and take its size to zero; the extra flux vanishes because $\mathbf{G}$ is regular there.

By translation and rotation invariance, use a radial Green function $G(R)$, where $R=|\mathbf{r}-{\mathbf{r}}^{\prime }|$ is the Euclidean distance. Away from the source, the radial modified Helmholtz equation isWriting $u=RG$ reduces this to $u^{\prime \prime }-{\kappa }^{2}u=0$, where $\kappa =|k|$. For $\kappa >0$, decay eliminates $e^{\kappa R}/R$, leaving $G=A{e}^{-\kappa R}/R$. For $\kappa =0$, the two radial branches are a constant and $1/R$; decay again leaves $A/R$.

To fix the singular normalization, integrate the defining equation over a ball of radius $\epsilon$ about ${\mathbf{r}}^{\prime }$. The divergence theorem givesThe volume term is $O({\epsilon }^2)$, whereasConsequentlySource qualification. The printed $e^{-kR}$ is the decaying branch only when $k\ge 0$, although the source declares $k$ real. For negative $k$ it grows, so it cannot be the requested decaying Green function. The absolute value above covers every real $k$ and reduces to the printed form for nonnegative $k$.

Take any constant vector $\mathbf{a}$. Its curl is zero, so part (a) and the divergence theorem givewhere $\mathbf{n}$ is the outward unit normal vector. The scalar triple product identity rewrites the last integrand as $\mathbf{a}\cdot (\mathbf{F}\times \mathbf{n})$. Since $\mathbf{a}$ is arbitrary, equality of these scalar products implies the vector identity

The same Maxwell equations givewhere the last equality uses $\nabla \cdot \mathbf{E}=0$. Combining with part (ii),The divergences of the symmetric gradients satisfyThus the local identity is ${\partial }_t(u_i^E+u_i^B)={\partial }_j(S_{ij}^E-c^2{S}_{ij}^B)$. Integrate over the fixed volume and apply the divergence theorem componentwise:Here $d{A}_j=n_{j}dA$ uses the outward normal vector.

The divergence of a cross product identity and the vacuum Maxwell equations implyThe divergence theorem therefore gives the required balance withAn additive time-independent constant in $\lambda$ is immaterial. This is the vacuum Poynting theorem in the normalization of the given equations: equivalently ${\partial }_t\lambda +\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. The order of the cross product accounts for the plus sign in the stated integral identity.

Use the sign convention in which a fundamental solution of the Laplace equation satisfiesin the sense of distributions. It is a harmonic function away from the source. Writing $\rho =|\mathbf{r}-{\mathbf{r}}^{\prime }|$, the radial equation is $(\rho G^{\prime }{)}^{\prime }=0$, so $G=A\log \rho +B$. The divergence theorem on a small circle about the source fixes its total flux:ConsequentlyThe arbitrary constant is a harmonic function. For the alternative convention $-\Delta G={\delta }^{(2)}$, the logarithm has the opposite sign; subsequent Green's function formulas must use that convention consistently.

Let $V\subset {\mathbb{R}}^3$ be a bounded region with piecewise smooth closed boundary $\partial V$, let $\mathbf{n}$ be its outward unit normal vector, and let $\mathbf{A}$ be a continuously differentiable vector field on a neighbourhood of its closure. The divergence theorem statesHere $dV$ is the positive volume measure, $dS$ is the surface area measure, $\nabla \cdot \mathbf{A}={\partial }_x{A}_x+{\partial }_y{A}_y+{\partial }_z{A}_z$, and the right side is the total outward flux. All boundary components must be included, with their normals pointing away from $V$.

For a nondegenerate surface take $R>0$, and abbreviate $C=\cosh R$, $H=\sinh R$. Close the upper oblate spheroid by its equatorial disk $D$, obtaining a half-ellipsoid $V$. The divergence isThe integral of $2y$ vanishes by reflection symmetry. At height $z$, the cross-section is a disk of radius $q=C\sqrt{1-z^2/H^2}$. Using planar polar coordinates,ConsequentlyThe outward unit normal vector on $D$ is $-{\mathbf{e}}_z$, so its flux is $-2\pi C^2$. Applying the divergence theorem to the closed surface and subtracting this disk contribution givesThis is a special case of polynomial flux through a half-ellipsoid. Including the closing disk is essential: the constant vertical component has zero divergence but nonzero flux through the open curved surface.

Suppose ${\Psi }_1$ and ${\Psi }_2$ have the same boundary data, and put $u={\Psi }_1-{\Psi }_2$. Then $\Delta u=0$ in $V$ and $u=0$ on its boundary. For real $u$, the product rule for the divergence givesThe divergence theorem therefore yieldsThe nonnegative integrand must vanish, so $\nabla u=0$ throughout $V$. On each connected component $u$ is constant, and its zero boundary value makes that constant zero. Hence $\boxed{{\Psi }_1={\Psi }_2}$, proving uniqueness. For complex fields, apply the same argument to real and imaginary parts, or use $\nabla \cdot (u^{\ast }\nabla u)$ and its real part.

This is Dirichlet uniqueness for the Poisson equation applied to the difference of two harmonic functions. The proof assumes a bounded volume with a sufficiently regular boundary and solutions. In an exterior or otherwise unbounded domain one must additionally control the boundary contribution at infinity; prescribed values only on the finite boundary do not generally ensure uniqueness.

Write $C=A\sqrt{\pi /(2k)}$, so the radial field is $\Phi =C{e}^{-kr}/r$. Integrate its equation over a ball of radius $ϵ$ centred at the source. The divergence theorem givesThe second term is $4\pi k^{2}C{\int }_0^{ϵ}r{e}^{-kr}dr=O({ϵ}^2)$ and vanishes in the limit. MeanwhileThusThe minus sign follows from the convention $(\Delta -k^2)\Phi ={\delta }^{(3)}$, opposite to the positive static Yukawa Green function for $-\Delta +k^2$. For arbitrary real $k$, replace it in the exponential by $|k|$ to select decay at infinity. The $k=0$ limit is $-1/(4\pi r)$.